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4 years ago

A sample of propane, C3H8, contains 13.8 moles of carbon atoms. How many total moles of atoms does the sample contain

Chemistry

1 answer:

aev [14]4 years ago

5 0

Answer:

$Total = 50.6\ moles$

Explanation:

Given

$Propane = C_3H_8$

Represent Carbon with C and Hydrogen with H

$C = 13.8$

Required

Determine the total moles

First, we need to represent propane as a ratio

$C_3H_8$ implies

$C:H = 3:8$

So, we're to first solve for H when $C = 13.8$

Substitute 13.8 for C

$13.8 : H = 3 : 8$

Convert to fraction

$\frac{13.8}{H} = \frac{3}{8}$

Cross Multiply

$3 * H = 13.8 * 8$

$3 H = 110.4$

Solve for H

$H = 110.4/3$

$H = 36.8$

So, when

$C = 13.8$

$H = 36.8$

$Total = C + H$

$Total = 13.8 + 36.8$

$Total = 50.6\ moles$

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Write a RACE paragraph to make and defend a claim about how salinity and temperature impact ocean density .

djverab [1.8K]

Answer:

Explanation:

Girl. No one is about to write a paragraph for you. Research, collect information and attempt it on your own. We are here to help, not to do your work for you.

7 0

4 years ago

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0

4 years ago

Aqueous aluminum bromide and solid zinc are produced by the reaction of solid aluminum and aqueous zinc bromide . write a balanc

Rashid [163]

3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s) The unbalanced equation is: ZnBr2 (aq) + Al (s) ==> AlBr3 (aq) +Zn (s) First, count the atoms of each element on each side of the equation: Zn 1,1 Br 2,3 Al 1,1 The zinc and aluminum, but the bromine doesn't match with 2 and 3. So look for the least common multiple of 2 and 3 which is 6 and adjust the quantities on both sides to have 6 bromine atoms on both sides. Do this by having 3 zinc bromide on the left and 2 aluminum bromide on the right, getting: 3 ZnBr2 (aq) + Al (s) ==> 2 AlBr3 (aq) +Zn (s) Now check the atom counts again for both sides: Zn 3,1 Br 6,6 Al 1,2 Now bromine matches, but zinc and aluminum doesn't. But it's easy enough to add an extra aluminum to the left and 2 more zinc to the right. Giving: 3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s) Now check the atom counts again: Zn 3,3 Br 6,6 Al 2,2 And they match. So the balanced equation is: 3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)

8 0

3 years ago

What volume (in L) will a 32 g sample of butane gas, C4H10(g), occupy at a temperature of 45.0 oC and a pressure of 728 mm Hg?

larisa86 [58]

Answer:

15.0 L

Explanation:

To find the volume, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (mmHg)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)

-----> T = temperature (K)

To calculate the volume, you need to (1) convert grams C₄H₁₀ to moles (via the molar mass), then (2) convert the temperature from Celsius to Kelvin, and then (3) calculate the volume (via the Ideal Gas Law).

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)

Molar Mass (C₄H₁₀): 58.124 g/mol

32 grams C₄H₁₀ 1 moles
------------------------- x ----------------------- = 0.551 moles C₄H₁₀
58.124 grams

P = 728 mmHg R = 62.36 L*mmHg/mol*K

V = ? L T = 45.0 °C + 273.15 = 318.15 K

n = 0.551 moles

PV = nRT

(728 mmHg)V = (0.551 moles)(62.36 L*mmHg/mol*K)(318.15 K)

(728 mmHg)V = 10922.7632

V = 15.0 L

6 0

2 years ago

Based on my previous question I have posted.. Answer this.. It's the continuation

Anna35 [415]

Answer:

We don't know what solvent X and solvent Y are, but from the chart, we can see that in solvent X, hydrochloric acid can conduct electricity (bulb lights up), and react with calcium carbonate.

So, we can say the electrical conductivity when HCl is dissolved in solvent X is high, and when HCl is dissolved in solvent Y, the electrical conductivity is low (because light bulb doesn't light up).

Additionally, in solvent X, HCl ionizes, this shows the property of acids: reacts with carbonates to give CO2 (because CO2 reacts with lime water to make it cloudy).

In solvent Y, HCl does not ionize, so there is no reaction between acid and calcium carbonate.

4 0

3 years ago